Chapter 12 problem 79

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Daniel Fernandez 3H
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Joined: Sat Jul 09, 2016 3:00 am

Chapter 12 problem 79

Postby Daniel Fernandez 3H » Fri Dec 02, 2016 8:57 pm

12.79
Calculate the pH of 0.15 m H 2 SO 4 (aq) at 25 C.

Emma_Green_2C
Posts: 21
Joined: Wed Sep 21, 2016 2:57 pm

Re: Chapter 12 problem 79

Postby Emma_Green_2C » Sat Dec 03, 2016 2:28 pm

Sulfuric acid is polyprotic so there are two deprotonizations. The first time, H2SO4 is completely ionized into HSO4- and H3O+. Since the ionization is complete, the concentration of HSO4- is 0.15M and the concentration of H3O+ is 0.15M (same concentration as H2SO4). The next ionization is incomplete, so an ICE table is needed. The equation for the second ionization is HSO4- + H2O (double arrow) SO42- + H3O+. The initial concentrations are taken from the 1st ionization, so [HSO4-]= 0.15M, [SO42-]=0, and [H3O+]=.15M. Set up the Ka as usual so it's (0.15+x)(x)/(0.15-x)= 1.2x10^-2. Use the second Ka value for H2SO4 which can be found on pg. 496 in the textbook. Ka is too large for x to be ignored, therefore use the quadratic formula to solve for x. X=0.0104M. 0.15+x is the equilibrium concentration of hydronium ions so the 0.15+0.0104=0.16M [H3O+]. Taking the negative log of that gives pH=0.80

Cobie_Allen_1H
Posts: 24
Joined: Wed Sep 21, 2016 2:56 pm

Re: Chapter 12 problem 79

Postby Cobie_Allen_1H » Sun Dec 04, 2016 11:11 am

Thank you! Your explanation is very helpful! I was also confused on how to do this problem!


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